Explanation

(d)
Given, $\left( \tan^{- 1}x \right)^{2} + \left( \cot^{- 1}x \right)^{2} = \frac{5\pi^{2}}{8}$
$\therefore\ \ \ \left( \tan^{- 1}{x + \cot^{- 1}x} \right)^{2} - 2\tan^{- 1}{x\left( \frac{\pi}{2} - \tan^{- 1}x \right)} = \frac{5\pi^{2}}{8}$
$\Rightarrow \ \ \ \ \ \ \ \ \frac{\pi^{2}}{4} - 2 \times \frac{\pi}{2} = \tan^{- 1}{x + 2}\left( \tan^{- 1}x \right)^{2}\frac{5\pi^{2}}{8}$
$\Rightarrow \ \ 2\left( \tan^{- 1}x \right)^{2} - \pi\tan^{- 1}{x - \frac{3\pi^{2}}{8} = 0}$
$\Rightarrow \ \tan^{- 1}{x = - \frac{\pi}{4},\ \frac{3\pi}{4}}$
Now, we take $\tan^{- 1}{x = - \frac{\pi}{4}\ \ \Rightarrow \ \ x = - 1}$